Скачать или смотреть [LeetCode 540]. Single Element in a Sorted Array - Optimal O(logN) Solution

Welcome to DevSkills! In this coding lesson, we break down LeetCode Problem #540: Single Element in a Sorted Array. This is a tricky problem that many developers struggle to analyze the critical index parity pattern.

🎯 Problem Overview & Required Complexity
We are given a sorted array in which every element appears exactly twice, except for a unique element. The strict constraints require the solution to run in O(logN) time and O(1) space. The O(logN) requirement is a strong hint that Binary Search must be used. We demonstrate why slower approaches, such as Linear Scan or XORing all elements (O(N) time), fail to meet the required complexity.

💡 Core Insight: The Index Parity Trick
Since we are constrained to O(logN) time, we must use binary search, but in a non-standard way—to find a transition point or breaking point, not just a value. The key is the array's index parity invariant:

• Before the single element, paired elements consistently start at an even index (even index,odd index).
• The single element shifts the pattern, causing subsequent pairs to start at an odd index (odd index, even index).

💻 The O(logN) Binary Search Algorithm

Our optimized algorithm leverages this parity shift:
1. Midpoint Adjustment: We enforce that the calculated mid index is always even (by decrementing if it is odd). This ensures nums[mid] is always the potential start of a pair.
2. Search Guidance: We compare nums[mid] with nums[mid + 1].
◦ If the pair is intact (nums[mid]==nums[mid+1]), the single element must be on the right side, and we safely discard the current pair: low=mid+2.
◦ If the pair is broken (nums[mid]==nums[mid+1]), the answer is at mid or to the left: high=mid.
3. Efficiency: This process repeatedly halves the search space, guaranteeing the required O(logN) time complexity.

We also cover an elegant XOR-based bit-manipulation approach (nums[mid] == nums[mid ^ 1]) that achieves the same result while simplifying the neighbor check.

⏱️ Complexity Analysis
• Time Complexity: O(logN) – Achieved because the binary search reduces the search space by half in every iteration.
• Space Complexity: O(1) – Only constant variables (low, high, mid) are used.
This method demonstrates "Interview-Level Reasoning," showing mastery of algorithmic constraints and array properties.

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